hdoj--2579--Dating with girls(2)（搜索+三维标记）

Dating with girls(2)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2691 Accepted Submission(s): 752



[align=left]Problem Description[/align]
If you have solved the problem Dating with girls(1).I think you can solve this problem too.This problem is also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze.If you can find
the girl, then you can date with the girl.Else the girl will date with other boys. What a pity!

The Maze is very strange. There are many stones in the maze. The stone will disappear at time t if t is a multiple of k(2<= k <= 10), on the other time , stones will be still there.

There are only ‘.’ or ‘#’, ’Y’, ’G’ on the map of the maze. ’.’ indicates the blank which you can move on, ‘#’ indicates stones. ’Y’ indicates the your location. ‘G’ indicates the girl's location . There is only one ‘Y’ and one ‘G’. Every seconds you can move
left, right, up or down.



[align=left]Input[/align]
The first line contain an integer T. Then T cases followed. Each case begins with three integers r and c (1 <= r , c <= 100), and k(2 <=k <= 10).

The next r line is the map’s description.

[align=left]Output[/align]
For each cases, if you can find the girl, output the least time in seconds, else output "Please give me another chance!".

[align=left]Sample Input[/align]

1
6 6 2
...Y..
...#..
.#....
...#..
...#..
..#G#.

[align=left]Sample Output[/align]

7

[align=left]Source[/align]
HDU 2009-5 Programming Contest

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
//三维标记 ，当前的时间与t取余，到达"#"的时间只有t种，
//如果这几种都不行，那这条路就废了
char map[110][110];
int a[110][110][12];
int m,n,t;
int sx,sy;
int dx[4]={1,0,0,-1};
int dy[4]={0,1,-1,0};
struct node
{
int x,y,step;
friend bool operator < (node s1 ,node s2)
{
return s1.step>s2.step;
}
}p,temp;
void bfs()
{
memset(a,0,sizeof(a));
queue<node>q;
p.x=sx,p.y=sy;
p.step=0;
q.push(p);
a[sx][sy][0]=1;
while(!q.empty())
{
p=q.front();
q.pop();
if(map[p.x][p.y]=='G')
{
printf("%d\n",p.step);
return ;
}
for(int i=0;i<4;i++)
{
temp.x=p.x+dx[i];
temp.y=p.y+dy[i];
temp.step=p.step+1;
int d=temp.step%t;
if(temp.x<0||temp.x>=n||temp.y<0||temp.y>=m||a[temp.x][temp.y][d])
continue;
a[temp.x][temp.y][d]=1;
if(map[temp.x][temp.y]=='#'&&d)
continue;
q.push(temp);
}
}
printf("Please give me another chance!\n");
}
int main()
{
int s;
scanf("%d",&s);
while(s--)
{
scanf("%d%d%d",&n,&m,&t);
for(int i=0;i<n;i++)
{
scanf("%s",map[i]);
for(int j=0;j<m;j++)
{
if(map[i][j]=='Y')
sx=i,sy=j;
}
}
bfs();
}
return 0;
}